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A cart is slowing down at a rate of
2t+60 centimeters per second per second (where
t is the time in seconds).
By how many centimeters per second does the cart slow down between
t=5 and
t=15 ?
Choose 1 answer:
(A) 20
(B) 75
(C) 800
(D) 875

A cart is slowing down at a rate of $2 t+60$ centimeters per second per second (where $t$ is the time in seconds). $\newline$ By how many centimeters per second does the cart slow down between $t=5$ and $t=15$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $20$ $\newline$ (B) $75$ $\newline$ (C) $800$ $\newline$ (D) $875$

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Solve equations with variables on both sides: word problems

Full solution

Q. A cart is slowing down at a rate of

2 t+60

centimeters per second per second (where

t

is the time in seconds).

\newline

By how many centimeters per second does the cart slow down between

t=5

and

t=15

?

\newline

Choose

1

answer:

\newline

(A)

20

\newline

(B)

75

\newline

(C)

800

\newline

(D)

875

Understand the problem: Understand the problem. $\newline$ We are given an acceleration function in terms of time, which is $2t + 60 \, \text{cm/s}^2$ . We need to find the change in velocity between $t=5$ and $t=15$ seconds.
Calculate acceleration at $t=5$ : Calculate the acceleration at $t=5$ seconds. Acceleration at $t=5$ is $a(5) = 2(5) + 60 = 10 + 60 = 70$ cm/s $^2$ .
Calculate acceleration at $t=15$ : Calculate the acceleration at $t=15$ seconds. Acceleration at $t=15$ is $a(15) = 2(15) + 60 = 30 + 60 = 90$ cm/s $^2$ .
Calculate average acceleration: Calculate the average acceleration between $t=5$ and $t=15$ seconds. $\newline$ The average acceleration is the sum of the accelerations at $t=5$ and $t=15$ divided by $2$ . $\newline$ Average acceleration = $(a(5) + a(15)) / 2 = (70 + 90) / 2 = 160 / 2 = 80$ cm/s $^2$ .
Calculate time interval: Calculate the time interval between $t=5$ and $t=15$ seconds. $\newline$ The time interval $\Delta t$ is $t_{\text{final}} - t_{\text{initial}} = 15 - 5 = 10$ seconds.
Calculate change in velocity: Calculate the change in velocity using the average acceleration and the time interval. Change in velocity $\Delta v = \text{average acceleration} \times \text{time interval} = 80 \, \text{cm/s}^2 \times 10 \, \text{s} = 800 \, \text{cm/s}$ .

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